CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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$x \mapsto x + 1/x$ on $[1,\infty)$ strictly decreases every distance and has no fixed point

Statement refuted

Refuted claim: a self-map $f$ of a nonempty complete metric space with $d(f(x),f(y)) < d(x,y)$ for all $x \ne y$ has a fixed point; equivalently, the contraction hypothesis of A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point may be weakened to that pointwise strict inequality (FALSE: $d(fx, fy) < d(x,y)$ for all $x \ne y$ on a complete metric space forces a fixed point).

Let $X := [1,\infty) \subseteq \mathbb{R}$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length) carry the metric $d(x,y) := |x-y|$ inherited from the real line (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset) and let

$f(x) := x + \frac{1}{x} \qquad (x \in X).$

Then $(X,d)$ is nonempty and complete, $f$ maps $X$ into $X$ , $d(f(x),f(y)) < d(x,y)$ whenever $x \ne y$ , and $f$ has no fixed point. Moreover $f$ is not a contraction (Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction): no real $q < 1$ satisfies $d(f(x),f(y)) \le q\,d(x,y)$ for all $x,y \in X$ .

Facts & Assumptions

Given: The interval $X = [1,\infty)$ with the metric $d$ inherited from $\mathbb{R}$ ; the map $f(x) = x + 1/x$ ; a natural $n \ge 1$ ; a real $q$ with $0 \le q < 1$ .

[L1]

The absolute value makes $\mathbb{R}$ a metric space, a restriction of a metric is a metric, and $|uv| = |u||v|$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset, Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Basic properties of the absolute value).

[L2]

$\mathbb{R}$ is complete; a closed subset of a complete metric space is complete; a subset is closed exactly when it is sequentially closed; and limits preserve non-strict inequalities ( $\mathbb{R}$ and $\mathbb{R}^n$ for $n \ge 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $\mathbb{R}$ , A subspace of a complete metric space is complete iff it is closed, and a complete subspace of any metric space is closed, A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed, Limits preserve non-strict inequalities, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$ , Complete metric space: every Cauchy sequence converges in the space).

[L3]

Positivity of inverses, reversal of order under reciprocation, and multiplication of inequalities by positives (Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication).

[L4]

Archimedean property and its reciprocal form: for every real $\eta > 0$ there is a natural $n \ge 1$ with $1/n < \eta$ ; and positive naturals sit in $\mathbb{R}$ in their own order (Every complete ordered field is Archimedean, For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Canonical naturals are positive and strictly increasing).

[L5]

Contraction: Lipschitz with a single constant $q$ satisfying $0 \le q < 1$ (Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction).

Counterexample

technique · direct

1.1

$X$ is nonempty ( $1 \in X$ ) and sequentially closed in $\mathbb{R}$ , since a sequence in $X$ converging to a real $p$ has $x_k \ge 1$ for every $k$ and hence $p \ge 1$ ; so $X$ is closed in $\mathbb{R}$ , and $\mathbb{R}$ being complete, $(X,d)$ is a nonempty complete metric space.

L1L2

1.2

$f$ maps $X$ into $X$ : $x \ge 1$ gives $1/x > 0$ , so $f(x) = x + 1/x > x \ge 1$ .

1.3

For $x,y \in X$ : $f(x) - f(y) = (x-y) + \big(1/x - 1/y\big) = (x-y) - \dfrac{x-y}{xy} = (x-y)\Big(1 - \dfrac{1}{xy}\Big)$ .

2.1

Let $x,y \in X$ with $x \ne y$ . Then $x,y \ge 1$ and they are not both $1$ , so $xy > 1$ , hence $0 < 1/(xy) < 1$ and $0 < 1 - 1/(xy) < 1$ ; therefore $d(f(x),f(y)) = |x-y| \cdot \big(1 - 1/(xy)\big) < |x-y| = d(x,y)$ .

step 1.2step 1.3L1L3

2.2

$f$ has no fixed point: $f(x) - x = 1/x > 0$ for every $x \in X$ , so $f(x) \ne x$ .

step 1.2L3

3.1

$f$ is not a contraction. Suppose $0 \le q < 1$ satisfied $d(f(x),f(y)) \le q\,d(x,y)$ for all $x,y \in X$ . Taking $x := n+1$ and $y := n$ for a natural $n \ge 1$ gives $|x - y| = 1$ and, by step 1.3, $d(f(x),f(y)) = 1 - \dfrac{1}{n(n+1)}$ , so $1 - \dfrac{1}{n(n+1)} \le q$ , that is $1 - q \le \dfrac{1}{n(n+1)} \le \dfrac{1}{n}$ .

step 1.3step 2.1L1L3L4L5

4.1

But $1 - q > 0$ , so [L4] supplies a natural $n \ge 1$ with $1/n < 1-q$ , contradicting step 3.1. Hence no such $q$ exists.

step 3.1L4

5.1

So $(X,d)$ is nonempty and complete, $f : X \to X$ strictly decreases every distance between distinct points, $f$ has no fixed point, and $f$ is not a contraction; this refutes the claim above and shows that the gap between the two hypotheses is real.

step 1.1step 1.2step 2.1step 2.2step 4.1∎

Remarks

The two hypotheses differ by a quantifier, and step 3.1 measures the gap. The shrinking factor at the pair $(n+1, n)$ is $1 - 1/(n(n+1))$ , which is below $1$ for every $n$ and approaches $1$ as $n$ grows; the contraction condition asks for a single bound below $1$ covering all pairs at once, and that is exactly what fails. The same quantifier move separates continuity from uniform continuity (Uniform continuity of a map of metric spaces: one $\delta$ serving every point).
Why the iterates do not help. Starting anywhere in $[1,\infty)$ , the iterates increase, since $f(x) > x$ , and they run off to the right; there is no Cauchy sequence to complete, so completeness of $X$ is no help at all. This is the opposite failure mode to $x \mapsto x/2$ maps $(0,1]$ into itself, is a $1/2$ -contraction, and has no fixed point, where the iterates are Cauchy and the space is missing their limit.
Unboundedness is essential to the example, not to the phenomenon as stated here. What this item establishes is only that completeness plus the strict inequality is not enough. It makes no claim about what additional hypothesis would suffice; the classical repair uses compactness, which is a later page of this library.

Depends on

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Nothing in the library uses this result yet.

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Sources

Contraction mapping (Wikipedia) (standard reference, not scraped)
Banach fixed-point theorem (Wikipedia) (standard reference, not scraped)

x↦x+1/xx \mapsto x + 1/xx↦x+1/x on [1,∞)[1,\infty)[1,∞) strictly decreases every distance and has no fixed point